제출 #762635

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
762635		GrindMachine	Salesman (IOI09_salesman)	C++17	60 / 100	601 ms	61460 KiB

salesman

// Om Namah Shivaya
 
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
 
#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)
 
template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}
 
template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}
 
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
 
/*
 
refs:
edi
 
*/
 
const int MOD = 1e9 + 7;
const int N = 5e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
 
template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051
 
    /*=======================================================*/
 
    struct data {
        ll a;
    };
 
    data neutral = {-inf2};
 
    data merge(data &left, data &right) {
        data curr;
        curr.a = max(left.a,right.a);
        return curr;
    }
 
    void create(int i, T v) {
        tr[i].a = v;
    }
 
    void modify(int i, T v) {
        amax(tr[i].a, v);
    }
 
    /*=======================================================*/
 
    int n;
    vector<data> tr;
 
    segtree() {
 
    }
 
    segtree(int siz) {
        init(siz);
    }
 
    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }
 
    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }
 
    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }
 
    data query(int l, int r) {
        data resl = neutral, resr = neutral;
 
        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) resl = merge(resl, tr[l++]);
            if (!(r & 1)) resr = merge(tr[r--], resr);
        }
 
        return merge(resl, resr);
    }
};
 
void solve(int test_case)
{
    ll n,U,D,s; cin >> n >> U >> D >> s;
    vector<array<ll,3>> a(n+5);
    rep1(i,n) rep(j,3) cin >> a[i][j];
 
    // sort(a.begin()+1, a.begin()+n+1);
 
    a[0] = {0,s,0};
    a[n+1] = {N-1,s,0};
        
    vector<int> here[N];
    rep1(i,n+1){
        here[a[i][0]].pb(i);
    }

    vector<ll> dp(n+5,-inf2);
    segtree<ll> st1(N), st2(N);
    
    dp[0] = 0;
    st1.pupd(s,s*D), st2.pupd(s,-s*U);
    
    rep1(t,N-1){
        trav(i,here[t]){
            auto [t,x,add] = a[i];
            ll v1 = st1.query(1,x).a - x*D + add;
            ll v2 = st2.query(x,N-1).a + x*U + add;
            dp[i] = max(v1,v2);
        }

        vector<array<ll,3>> order;
        trav(i,here[t]){
            order.pb({a[i][1],a[i][2],i});
        }

        sort(all(order));

        ll mx = -inf2;

        for(auto [pos,add,i] : order){
            amax(dp[i], mx);
            amax(mx, dp[i]-add);
            mx += add;
        }

        reverse(all(order));
        mx = -inf2;
    
        for(auto [pos,add,i] : order){
            amax(dp[i],mx);
            amax(mx,dp[i]-add);
            mx += add;
        }

        trav(i,here[t]){
            auto [t,x,add] = a[i];
            st1.pupd(x,dp[i]+x*D);
            st2.pupd(x,dp[i]-x*U);
        }
    }

    ll ans = dp[n+1];
    cout << ans << endl;
}
 
int main()
{
    fastio;
 
    int t = 1;
    // cin >> t;
 
    rep1(i, t) {
        solve(i);
    }
 
    return 0;
}

• Subtask 1: 60 / 100